In the sequence S of numbers

[jigar24]

In the sequence S of numbers, each term after the
first two terms is the sum of the two immediately
preceding terms. What is the 5th term of S ?
(1) The 6th term of S minus the 4th term equals 5.
(2) The 6th term of S plus the 7th term equals 21.

Source: GMAT Prep free tests

I am could solve the fist part -

4th term + 5th = 6th term
5th term = 6th term - 4th term
= 5. Hence A sufficient. Struggling with analyzing statement 2. Help please.

Correct answer A

[RonPurewal]

Struggling with analyzing statement 2. Help please.

if you're struggling with statement 2 here, i'd like to humbly suggest that you're still struggling with the basic format/definitions of data sufficiency in general, rather than with this problem in particular.

what i mean is this:
"sufficient" here means there's only one possible value for the 5th term.
"not sufficient" means the 5th term has more than one possible value.

^^^ if you realize that your goal is to figure out which of these is true, then you shouldn't have a very hard time with statement 2.
after all, the only thing you know in statement 2 is "6th + 7th = 21".
this means that the 6th and 7th terms can be any two numbers at all that add to 21.
once you've got that, you know that the 5th term is the 7th term minus the 6th term (because 5th + 6th = 7th)... but that's going to turn out in lots of different ways, because there are so many different possibilities for the 6th and 7th terms.
for instance, if the 6th term is 10 and the 7th term is 11, then the 5th term is 1.
on the other hand, if the 6th term is 1 and the 7th term is 20, then the 5th term is 19.
boom, insufficient.

this isn't really the kind of problem on which algebra is your friend, but you can even do that, too.
if the 6th term is "x", then the 7th term is (21 - x). in that case, the 5th term is (21 - x) - x = 21 - 2x, a value that's definitely going to depend on x (and so will have lots of different possible values).

again, if you have any trouble with statement 2 here, you should ...
... (i) go back to the basics of DS and make sure you FULLY understand the concepts of "sufficient" and "not sufficient"
... (ii) always define exactly what "sufficient" and "not sufficient" mean on any particular problem before you actually start doing math/work
... (iii) be able to TEST CASES on DS problems (as in the first approach above). if you are just trying to use algebra on every single problem, you'll find DS much harder than it really should be.